Areas Related To CirclesClass 10 Mathematics NCERT Solutions

Given: Radius of the circle, $r = 6 \text{ cm}$ Angle of the sector, $\theta = 60^{\circ}$

To Find: The area of the sector.

Formula: Area of a sector of angle $\theta$ is given by: $A = \frac{\theta}{360^{\circ}} \times \pi r^2$

Solution: Substituting the given values into the formula: $A = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (6)^2$ $A = \frac{1}{6} \times \frac{22}{7} \times 36$ $A = 1 \times \frac{22}{7} \times 6$ $A = \frac{132}{7} \text{ cm}^2$ $A \approx 18.86 \text{ cm}^2$

Final Answer: The area of the sector is $\frac{132}{7} \text{ cm}^2$ or approximately $18.86 \text{ cm}^2$.

Given: Circumference of the circle = 22 cm. A quadrant of a circle has a central angle of $90^{\circ}$.

To Find: The area of the quadrant.

Formula: Circumference of a circle, $C = 2 \pi r$ Area of a quadrant = $\frac{1}{4} \pi r^2$

Solution: First, we find the radius of the circle from its circumference. $2 \pi r = 22$ $2 \times \frac{22}{7} \times r = 22$ $r = \frac{22 \times 7}{2 \times 22}$ $r = \frac{7}{2} \text{ cm}$

Now, we calculate the area of the quadrant. Area of quadrant = $\frac{1}{4} \pi r^2$ $= \frac{1}{4} \times \frac{22}{7} \times (\frac{7}{2})^2$ $= \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$ $= \frac{1 \times 22 \times 7}{4 \times 2 \times 2}$ $= \frac{154}{16}$ $= \frac{77}{8} \text{ cm}^2$ $= 9.625 \text{ cm}^2$

Final Answer: The area of the quadrant is $\frac{77}{8} \text{ cm}^2$ or $9.625 \text{ cm}^2$.

Given: Length of the minute hand (radius), $r = 14 \text{ cm}$ Time duration = 5 minutes.

To Find: The area swept by the minute hand in 5 minutes.

Solution: The minute hand of a clock completes a full circle ($360^{\circ}$) in 60 minutes. Angle swept in 60 minutes = $360^{\circ}$ Angle swept in 1 minute = $\frac{360^{\circ}}{60} = 6^{\circ}$ Angle swept in 5 minutes, $\theta = 5 \times 6^{\circ} = 30^{\circ}$

The area swept is the area of a sector with angle $30^{\circ}$ and radius 14 cm.

Formula: Area of a sector, $A = \frac{\theta}{360^{\circ}} \times \pi r^2$

Calculation: $A = \frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (14)^2$ $A = \frac{1}{12} \times \frac{22}{7} \times 14 \times 14$ $A = \frac{1}{12} \times 22 \times 2 \times 14$ $A = \frac{1}{3} \times 11 \times 14$ $A = \frac{154}{3} \text{ cm}^2$ $A \approx 51.33 \text{ cm}^2$

Final Answer: The area swept by the minute hand in 5 minutes is $\frac{154}{3} \text{ cm}^2$ or approximately $51.33 \text{ cm}^2$.

Given: Radius of the circle, $r = 15 \text{ cm}$ Angle at the centre, $\theta = 60^{\circ}$ Use $\pi = 3.14$ and $\sqrt{3} = 1.73$

To Find: Areas of the minor and major segments.

Formula: Area of minor segment = Area of minor sector - Area of triangle

First, calculate the area of the minor sector. Area of minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2$ $= \frac{60^{\circ}}{360^{\circ}} \times 3.14 \times (15)^2$ $= \frac{1}{6} \times 3.14 \times 225$ $= 117.75 \text{ cm}^2$

Next, calculate the area of the triangle formed by the chord and two radii. Since the angle at the centre is $60^{\circ}$ and the two sides are radii (equal), the triangle is equilateral with side length 15 cm. Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times (\text{side})^2$ $= \frac{1.73}{4} \times (15)^2$ $= \frac{1.73}{4} \times 225$ $= 97.3125 \text{ cm}^2$

Area of minor segment = $117.75 - 97.3125 = 20.4375 \text{ cm}^2$

Formula: Area of major segment = Area of circle - Area of minor segment

First, calculate the area of the circle. Area of circle = $\pi r^2 = 3.14 \times (15)^2 = 3.14 \times 225 = 706.5 \text{ cm}^2$

Area of major segment = $706.5 - 20.4375 = 686.0625 \text{ cm}^2$

Final Answer: The area of the minor segment is $20.4375 \text{ cm}^2$. The area of the major segment is $686.0625 \text{ cm}^2$.

Given: Radius of the circle, $r = 12 \text{ cm}$ Angle at the centre, $\theta = 120^{\circ}$ Use $\pi = 3.14$ and $\sqrt{3} = 1.73$

To Find: Area of the corresponding segment (minor segment).

Formula: Area of segment = Area of sector - Area of triangle

First, calculate the area of the sector OAB. Area of sector = $\frac{\theta}{360^{\circ}} \times \pi r^2$ $= \frac{120^{\circ}}{360^{\circ}} \times 3.14 \times (12)^2$ $= \frac{1}{3} \times 3.14 \times 144$ $= 3.14 \times 48$ $= 150.72 \text{ cm}^2$

Next, calculate the area of $\triangle OAB$. To find the area of $\triangle OAB$, draw a perpendicular OM from O to the chord AB. This perpendicular bisects the chord AB and the angle $\angle AOB$. So, $\angle AOM = \angle BOM = \frac{120^{\circ}}{2} = 60^{\circ}$. In the right-angled triangle $\triangle OMA$: $OM = OA \cos(60^{\circ}) = 12 \times \frac{1}{2} = 6 \text{ cm}$ $AM = OA \sin(60^{\circ}) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm}$ Length of the chord $AB = 2 \times AM = 2 \times 6\sqrt{3} = 12\sqrt{3} \text{ cm}$

Area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OM$ $= \frac{1}{2} \times 12\sqrt{3} \times 6$ $= 36\sqrt{3} \text{ cm}^2$ Using $\sqrt{3} = 1.73$: Area of $\triangle OAB = 36 \times 1.73 = 62.28 \text{ cm}^2$

Now, calculate the area of the segment. Area of segment = Area of sector - Area of $\triangle OAB$ $= 150.72 - 62.28$ $= 88.44 \text{ cm}^2$

Final Answer: The area of the corresponding segment is $88.44 \text{ cm}^2$.

Given: Number of ribs = 8. The ribs are equally spaced. Radius of the flat circle, $r = 45 \text{ cm}$.

To Find: The area between two consecutive ribs.

Solution: The area between two consecutive ribs is the area of one sector. Since there are 8 equally spaced ribs, the circle is divided into 8 equal sectors.

Angle of each sector, $\theta = \frac{360^{\circ}}{8} = 45^{\circ}$.

Formula: Area of a sector = \frac{\theta}{360^{\circ}}} \times \pi r^2

Calculation: Area = $\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (45)^2$ $= \frac{1}{8} \times \frac{22}{7} \times 45 \times 45$ $= \frac{11}{4} \times \frac{1}{7} \times 2025$ $= \frac{22275}{28} \text{ cm}^2$ $\approx 795.536 \text{ cm}^2$

Final Answer: The area between the two consecutive ribs of the umbrella is $\frac{22275}{28} \text{ cm}^2$ or approximately $795.54 \text{ cm}^2$.

Given: Number of wipers = 2. Length of each blade (radius), $r = 25 \text{ cm}$. Angle of sweep, $\theta = 115^{\circ}$. The wipers do not overlap.

To Find: The total area cleaned at each sweep of the blades.

Solution: The area cleaned by one wiper is the area of a sector.

Formula: Area of a sector = $\frac{\theta}{360^{\circ}} \times \pi r^2$

Calculation for one wiper: Area cleaned by one wiper = $\frac{115^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (25)^2$ $= \frac{23}{72} \times \frac{22}{7} \times 625$ $= \frac{23}{36} \times \frac{11}{7} \times 625$ $= \frac{158125}{252} \text{ cm}^2$

Since there are two identical wipers, the total area cleaned is twice the area cleaned by one wiper. Total area cleaned = $2 \times \frac{158125}{252}$ $= \frac{158125}{126} \text{ cm}^2$ $\approx 1254.96 \text{ cm}^2$

Final Answer: The total area cleaned at each sweep of the blades is $\frac{158125}{126} \text{ cm}^2$ or approximately $1254.96 \text{ cm}^2$.

Given: Angle of the sector, $\theta = 80^{\circ}$. Distance the light spreads (radius), $r = 16.5 \text{ km}$. Use $\pi = 3.14$.

To Find: The area of the sea over which the ships are warned.

Solution: The area is that of a sector of a circle.

Formula: Area of a sector = $\frac{\theta}{360^{\circ}} \times \pi r^2$

Calculation: Area = $\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times (16.5)^2$ $= \frac{2}{9} \times 3.14 \times 16.5 \times 16.5$ $= \frac{2}{9} \times 3.14 \times 272.25$ $= \frac{1709.73}{9}$ $= 189.97 \text{ km}^2$

Final Answer: The area of the sea over which the ships are warned is $189.97 \text{ km}^2$.

Given: Number of equal designs = 6. Radius of the cover, $r = 28 \text{ cm}$. Cost rate = ₹ 0.35 per $\text{cm}^2$. Use $\sqrt{3} = 1.7$.

To Find: The total cost of making the designs.

Solution: The 6 equal designs are 6 equal segments of the circle. The angle of the sector corresponding to each design at the center is $\theta = \frac{360^{\circ}}{6} = 60^{\circ}$.

First, find the area of one design (one segment). Area of one segment = Area of sector - Area of triangle.

Area of the sector: $= \frac{60^{\circ}}{360^{\circ}} \times \pi r^2$ $= \frac{1}{6} \times \frac{22}{7} \times (28)^2$ $= \frac{1}{6} \times \frac{22}{7} \times 28 \times 28$ $= \frac{1}{6} \times 22 \times 4 \times 28$ $= \frac{2464}{6} = \frac{1232}{3} \approx 410.67 \text{ cm}^2$

Area of the triangle: Since the angle at the center is $60^{\circ}$ and the two sides are radii, the triangle is equilateral with side length 28 cm. Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times (\text{side})^2$ $= \frac{1.7}{4} \times (28)^2$ $= \frac{1.7}{4} \times 784$ $= 1.7 \times 196 = 333.2 \text{ cm}^2$

Area of one design = Area of sector - Area of triangle $= 410.67 - 333.2 = 77.47 \text{ cm}^2$

Total area of 6 designs = $6 \times$ Area of one design $= 6 \times 77.47 = 464.82 \text{ cm}^2$

Total cost of making the designs = Total area $\times$ Cost rate $= 464.82 \times 0.35$ $= 162.687$

Final Answer: The cost of making the designs is ₹ 162.69 (rounded to two decimal places).

Given: Angle of the sector = $p$ (in degrees) Radius of the circle = R

To Find: The correct formula for the area of the sector.

Formula: The standard formula for the area of a sector with angle $\theta$ and radius $r$ is: $\text{Area} = \frac{\theta}{360} \times \pi r^2$

Solution: Substitute $\theta = p$ and $r = R$ into the standard formula: $\text{Area} = \frac{p}{360} \times \pi R^2$

Now, we check the given options: (A) $\frac{p}{180} \times 2 \pi \mathrm{R}$ : This is a formula for arc length with an incorrect denominator. (B) $\frac{p}{180} \times \pi \mathrm{R}^{2}$ : This is a formula for area with an incorrect denominator. (C) $\frac{p}{360} \times 2 \pi \mathrm{R}$ : This is the formula for the length of an arc. (D) $\frac{p}{720} \times 2 \pi \mathrm{R}^{2}$ : Let's simplify this expression. $\frac{p}{720} \times 2 \pi \mathrm{R}^{2} = \frac{p \times 2}{720} \times \pi R^2 = \frac{p}{360} \times \pi R^2$ This matches the standard formula for the area of a sector.

Final Answer: The correct answer is (D).